Question

proof of question no 71

Answer 1

Consider LHS:
sin6A+cos6A = (sin2A)3 + (cos2A)3
                     = (sin2A + cos2A)3 − 3 sin2Acos2A(sin2A + cos2A)  [Since, (a+b)3 = a3 + b3+3ab(a+b)]
                     = (1)3 − 3 sin2A cos2A(1)  [Since, sin2A + cos2A = 1]
                     =   1 − 3 sin2A cos2A
                     = RHS

plzz mark as brainliest answer !!!

Answer 2

To prove sin^6A + cos^6A = 1 - 3sin^2Acos^2A
LHS
sin^6 A + cos^6 A = (sin^2 A )^3 + (cos^2 A)
= ( sin^2 A + cos^ A )^3 - 3sin^Acos^A(sin^A + cos^A)
= ( 1 ) - 3sin^2cos^2A(1)
= 1 - 3sin^Acos^A
= RHS
hence proved